통신이론Ch03 by KPC Korea

Chapter 3

Basic Modulation Techniques 3.1

Problems

Problem 3.1 The demodulated output, in general, is yD (t) = Lp{xc (t) 2 cos[ω c t + θ (t)]} where Lp {•} denotes the lowpass portion of the argument. With xc (t) = Ac m (t) cos [ω c t + φ0 ] the demodulated output becomes yD (t) = Lp {2Ac m (t) cos [ω c t + φ0 ] cos [ω c t + θ (t)]} Performing the indicated multiplication and taking the lowpass portion yields yD (t) = Ac m (t) cos [θ (t) − φ0 ] If θ(t) = θ0 (a constant), the demodulated output becomes yD (t) = Ac m (t) cos [θ0 − φ0 ] Letting Ac = 1 gives the error ε (t) = m (t) [1 − cos (θ0 − φ0 )] The mean-square error is E 2 ® D 2 ε (t) = m (t) [1 − cos (θ0 − φ0 )]2 1

CHAPTER 3. BASIC MODULATION TECHNIQUES

where h·i denotes the time-average value. Since the term [1 − cos (θ0 − φ0 )] is a constant, we have 2 ® 2 ® ε (t) = m (t) [1 − cos (θ0 − φ0 )]2 Note that for θ0 = φ0 , the demodulation carrier is phase coherent with the original modulation carrier, and the error is zero. For θ (t) = ω0 t we have the demodulated output yD (t) = Ac m (t) cos (ω0 t − φ0 ) Letting Ac = 1, for convenience, gives the error ε (t) = m (t) [1 − cos (ω 0 t − φ0 )] giving the mean-square error E 2 ® D 2 ε (t) = m (t) [1 − cos (ω 0 t − φ0 )]2

In many cases, the average of a product is the product of the averages. (We will say more about this in Chapters 4 and 5). For this case E 2 ® 2 ®D ε (t) = m (t) [1 − cos (ω 0 t − φ0 )]2

Note that 1 − cos (ω 0 t − φ0 ) is periodic. Taking the average over an integer number of periods yields E D ® = 1 − 2 cos (ω 0 t − φ0 ) + cos2 (ω 0 t − φ0 ) [1 − cos (ω 0 t − φ0 )]2 = 1+

Thus

3 1 = 2 2

2 ® 3 2 ® m (t) ε (t) = 2

Problem 3.2 Multiplying the AM signal xc (t) = Ac [1 + amn (t)] cos ω c t by xc (t) = Ac [1 + amn (t)] cos ω c t and lowpass Þltering to remove the double frequency (2ω c ) term yields yD (t) = Ac [1 + amn (t)] cos θ (t)

3.1. PROBLEMS

xC ( t ) yD ( t )

Figure 3.1: For negligible demodulation phase error, θ(t) ≈ 0, this becomes yD (t) = Ac + Ac amn (t) The dc component can be removed resulting in Ac amn (t), which is a signal proportional to the message, m (t). This process is not generally used in AM since the reason for using AM is to avoid the necessity for coherent demodulation. Problem 3.3 A full-wave rectiÞer takes the form shown in Figure 3.1. The waveforms are shown in Figure 3.2, with the half-wave rectiÞer on top and the full-wave rectiÞer on the bottom. The message signal is the envelopes. Decreasing exponentials can be drawn from the peaks of the waveform as depicted in Figure 3.3(b) in the text. It is clear that the full-wave rectiÞed xc (t) deÞnes the message better than the half-wave rectiÞed xc (t) since the carrier frequency is eﬀectively doubled. Problem 3.4 Part a b c

2 ® mn (t) 1/3 1/3 1

a = 0.4 Ef f = 5, 1% Ef f = 5.1% Ef f = 13.8%

a = 0.6 Ef f = 10.7% Ef f = 10.7% Ef f = 26.5%

a=1 Ef f = 25% Ef f = 25% Ef f = 50%

Problem 3.5 By inspection, the normalized message signal is as shown in Figure 3.3. Thus 2 T 0≤t≤ mn (t) = t, T 2

CHAPTER 3. BASIC MODULATION TECHNIQUES

2 1.5 1 0.5 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

2 1.5 1 0.5 0

Figure 3.2:

mn ( t )

1 t

0 T -1

Figure 3.3:

3.1. PROBLEMS and

Also

2 ® 2 mn (t) = T

T /2 µ

¶ µ ¶ µ ¶ 2 2 2 2 21 T 3 1 = t dt = T T T 3 2 3

Ac [1 + a] = 40 Ac [1 − a] = 10 This yields

1+a 40 = =4 1−a 10

1 + a = 4 − 4a 5a = 3 Thus a = 0.6 Since the index is 0.6, we can write Ac [1 + 0.6] = 40 This gives Ac =

40 = 25 1.6

This carrier power is 1 1 Pc = A2c = (25)2 = 312.5 Watts 2 2 The eﬃciency is Ef f = Thus

(0.6)2

¡1¢

3 ¡ ¢ + (0.6)2 13

0.36 = 0.107 = 10.7% 3.36

Psb = 0.107 Pc + Psb where Psb represents the power in the sidebands and Pc represents the power in the carrier. The above expression can be written Psb = 0.107 + 0.107Psb This gives Psb =

0.107 Pc = 97.48 Watts 1.0 − 0.107

CHAPTER 3. BASIC MODULATION TECHNIQUES

Problem 3.6 For the Þrst signal τ=

T , 6

mn (t) = m (t)

and Ef f = 81.25%

and for the second signal τ=

5T , 6

1 mn (t) = m (t) 5

and Ef f = 14.77%

Problem 3.7 (a) The Þrst step in the solution to this problem is to plot m (t), or use a root-Þnding algorithm in order to determine the minimum value of m (t). We Þnd that the minimum of m (t) = −11.9523, and the minimum falls at t = 0.0352 and t = 0.0648. Thus the normalized message signal is mn (t) =

1 [9 cos 20πt − 7 cos 60πt] 11.9523

With the given value of c (t) and the index a, we have xc (t) = 100 [1 + 0.5mn (t)] cos 200πt This yields xc (t) = −14.6415 cos 140πt + 18.8428 cos 180πt +100 cos 200πt +18.8248 cos 220πt − 14.6415 cos 260πt We will need thislater to® plot the spectrum. (b) The value of m2n (t) is 2 ® mn (t) =

1 11.9523

¶2 µ ¶ h i 1 (9)2 + (7)2 = 0.455 2

(c)This gives the eﬃciency

(0.5)2 (0.455) = 10.213% 1 + (0.5)2 (0.455)

3.1. PROBLEMS

A f

-130 –110 –100 –90 -70

70 90 100 110 130

Figure 3.4: (d) The two-sided amplitude spectrum is shown in Figure 3.4. where 14.4615 = 7.2307 A= 2 and 18.8248 = 9.4124 B= 2 The phase spectrum results by noting that A is negative and all other terms are positive. (e) By inspection the signal is of the form xc (t) = a(t)c(t) where a(t) is lowpass and c(t( is highpass. The spectra of a(t) and c(t) are do not overlap. Thus the Hilbert transform of c(t) is c(t) x bc (t) = a(t)b

in which c(t) = 100 cos 200πt. This the envelope is q e(t) = 100 a2 (t) cos2 200πt + a2 (t) sin2 200πt = 100a(t)

where

a(t) = [1 + 0.5mn (t)]

Problem 3.8

CHAPTER 3. BASIC MODULATION TECHNIQUES

1 (a) mn (t) = 16 [9 cos 20πt + 7 cos 60πt] i 2 ® ¡ 1 ¢2 ¡ 1 ¢ h 2 2 = 0.2539 (b) mn (t) = 16 + (7) (9) 2

0.25(0.2539) = 0.05969 = 5.969% (c) Ef f = 1+0.25(0.2539) (d) The expression for xc (t) is · ¸ 1 (9 cos 20πt + cos 60πt) cos 200πt xc (t) = 100 1 + 32 = 10.9375 cos 140πt + 14.0625 cos 180πt

+100 cos 200πt +14.0625 cos 220πt + 10.9375 cos 260πt Note that all terms are positive so the phase spectrum is everywhere zero. The amplitude spectrum is identical to that shown in the previous problem except that A = B =

1 (10.9375) = 5.46875 2 1 (14.0625) = 7.03125 2

(e) As in the previous problem, the signal is of the form xc (t) = a(t)c(t) where a(t) is lowpass and c(t( is highpass. The spectra of a(t) and c(t) are do not overlap. Thus the Hilbert transform of c(t) is c(t) x bc (t) = a(t)b

where c(t) = 100 cos 200πt. This the envelope is q e(t) = 100 a2 (t) cos2 200πt + a2 (t) sin2 200πt ¸ · 1 (9 cos 20πt + cos 60πt) = 100a(t) = 100 1 + 32 Problem 3.9 The modulator output xc (t) = 40 cos 2π (200) t + 4 cos 2π (180) t + 4 cos 2π (220) t can be written xc (t) = [40 + 8 cos 2π (20) t] cos 2π (200) t

3.1. PROBLEMS

· ¸ 8 cos 2π (20) t cos 2π (200) t xc (t) = 40 1 + 40

By inspection, the modulation index is a=

8 = 0.2 40

Since the component at 200 Hertz represents the carrier, the carrier power is Pc =

1 (40)2 = 800 Watts 2

The components at 180 and 220 Hertz are sideband terms. Thus the sideband power is Psb =

1 1 (4)2 + (4)2 = 16 Watts 2 2

Thus, the eﬃciency is Ef f =

Psb 16 = = 0.0196 = 1.96% Pc + Psb 800 + 16

Problem 3.10 A = 14.14

B = 8.16

a = 1.1547

Problem 3.11 The modulator output xc (t) = 20 cos 2π (150) t + 6 cos 2π (160) t + 6 cos 2π (140) t is

· ¸ 12 cos 2π (10) t cos 2π (150) t xc (t) = 20 1 + 20

Thus, the modulation index, a, is a=

12 = 0.6 20

The carrier power is Pc =

1 (20)2 = 200 Watts 2

CHAPTER 3. BASIC MODULATION TECHNIQUES

and the sideband power is Psb =

1 1 (6)2 + (6)2 = 36 Watts 2 2

Thus, the eﬃciency is Ef f =

36 = 0.1525 200 + 36

Problem 3.12 (a) By plotting m (t) or by using a root-Þnding algorithm we see that the minimum value of m (t) is M = −3.432. Thus mn (t) = 0.5828 cos (2πfm t) + 0.2914 cos (4πfm t) + 0.5828 cos (10πfm t) The AM signal is xc (t) = Ac [1 + 0.7mn (t)] cos 2πfc t = 0.2040Ac cos 2π (fc − 5fm ) t

+0.1020Ac cos 2π (fc − 2fm ) t +0.2040Ac cos 2π (fc − fm ) t

+Ac cos 2πfc t +0.2040Ac cos 2π (fc + fm ) t +0.1020Ac cos 2π (fc + 2fm ) t +0.2040Ac cos 2π (fc + 5fm ) t The spectrum is drawn from the expression for xc (t). It contains 14 discrete components as shown Comp 1 2 3 4 5 6 7

Freq −fc − 5fm −fc − 2fm −fc − fm −fc −fc + fm −fc + 2fm −fc + 5fm

(b) The eﬃciency is 15.8%. Problem 3.13

Amp 0.102Ac 0.051Ac 0.102Ac 0.5Ac 0.102Ac 0.051Ac 0.102Ac

Comp 8 9 10 11 12 13 14

Freq fc − 5fm fc − 2fm fc − fm fc fc + fm fc + 2fm fc + 5fm

Amp 0.102Ac 0.051Ac 0.102Ac 0.5Ac 0.102Ac 0.051Ac 0.102Ac

3.1. PROBLEMS

ℑ{m ( t )}

{

Filter Characteristic

}

ℑ m2 (t )

fc − W

fc + W

2 fc

Figure 3.5: (a) From Figure 3.75 x (t) = m (t) + cos ω c t With the given relationship between x (t) and y (t) we can write y (t) = 4 {m (t) + cos ω c t} + 10 {m (t) + cos ω c t}2 which can be written y (t) = 4m (t) + 4 cos ω c t + 10m2 (t) + 20m (t) cos ω c t + 5 + 5 cos 2ω c t The equation for y (t) is more conveniently expressed y (t) = 5 + 4m (t) + 10m2 (t) + 4 [1 + 5m (t)] cos ω c t + 5 cos 2ωc t (b) The spectrum illustrating the terms involved in y (t) is shown in Figure 3.5. The center frequency of the Þlter is fc and the bandwidth must be greater than or equal to 2W . In addition, fc − W > 2W or fc > 3W , and fc + W < 2fc . The last inequality states that fc > W , which is redundant since we know that fc > 3W . (c) From the deÞnition of m (t) we have m (t) = Mmn (t) so that g (t) = 4 [1 + 5Mmn (t)] cos ω c t It follows that a = 0.8 = 5M

CHAPTER 3. BASIC MODULATION TECHNIQUES

Thus

0.8 = 0.16 5 (d) This method of forming a DSB signal avoids the need for a multiplier. M=

Problem 3.14 1 sgn (f + fc ) 2 1 1 xU SB (t) = Ac m (t) cos ω c t − Ac m b (t) sin ω c t 2 2 HU (f ) = 1 −

Problem 3.15 For the USB SSB case, the modulator output is a sinusoid of frequency fc + fm , while for the LSB SSB case, the modulator output is a sinusoid of frequency of fc − fm . Problem 3.16 Using Figure 3.13 and the phases given in the problem statement, the modulator output becomes xc (t) =

Aε cos [(ω c − ω1 ) t + φ] 2 A (1 − ε) cos [(ω c + ω1 ) t + θ 1 ] + 2 B + cos [(ω c + ω 2 ) t + θ2 ] 2

Multiplying xc (t) by 4 cos ω c t and lowpass Þltering yields the demodulator output yD (t) = Aε cos (ω 1 − φ) + A (1 − ε) cos (ω 1 t + θ1 ) + B cos (ω 2 t + θ2 ) For the sum of the Þrst two terms to equal the desired output with perhaps a time delay, θ1 must equal−φ. This gives yD (t) = A cos (ω 1 t + φ) + B cos (ω 2 t + θ2 ) which we can write

µ ¶ µ ¶ θ1 θ2 + B cos ω 2 t + yD (t) = A cos ω1 t + ω1 ω2

For no distortion, yD (t) must be of the form m (t − τ ). Thus θ1 θ2 = ω1 ω2

3.1. PROBLEMS

so that

ω2 θ1 ω1 Thus, the phase must be linear. The fact that φ = −θ1 tells us that the Þlter phase response must have odd symmetry about the carrier frequency. θ2 =

Problem 3.17 We assume that the VSB waveform is given by xc (t) =

1 Aε cos (ω c − ω 1 ) t 2 1 + A (1 − ε) cos (ω c + ω 1 ) t 2 1 + B cos (ω c + ω 2 ) t 2

We let y (t) be xc (t) plus a carrier. Thus y (t) = xc (t) + K cos ω c t It can be shown that y (t) can be written y (t) = y1 (t) cos ω c (t) + y2 (t) sin ω c t where A B cos ω 1 t + cos ω 2 t + K 2 µ ¶ 2 A B Aε + sin ω 1 t − sin ω 2 t y2 (t) = 2 2

y1 (t) =

Also y (t) = R (t) cos (ω c t + θ) where R (t) is the envelope and is therefore the output of an envelope detector. It follows that q R (t) = y12 (t) + y22 (t)

For K large, R (t) = |y1 (t)|, which is 12 m (t) + K, where K is a dc bias. Thus if the detector is ac coupled, K is removed and the output y (t) is m (t) scaled by 12 . Problem 3.18 The required Þgure appears in Figure 3.6. Problem 3.19

CHAPTER 3. BASIC MODULATION TECHNIQUES

Desired Signal

ω1 Local Oscillator

ω1 − ω 2 Signal at Mixer Output

ω 2 − ω IF

2ω1 − ω 2

Image Signal

ω 2ω1 − ω 2 Image Signal at Mixer Output

ω 3ω 2 − 2ω1

ω2

Figure 3.6:

3.1. PROBLEMS

Since high-side tuning is used, the local oscillator frequency is fLO = fi + fIF where fi , the carrier frequency of the input signal, varies between 5 and 25 M Hz. The ratio is fIF + 25 R= fIF + 5 where fIF is the IF frequency expressed in MHz. We make the following table fIF ,MHz 0.4 0.5 0.7 1.0 1.5 2.0

R 4.70 4.63 4.51 4.33 4.08 3.86

A plot of R as a function of fIF is the required plot. Problem 3.20 For high-side tuning we have fLO = fi + fIF = 1120 + 455 = 1575 kHz fIM AGE = fi + 2fIF = 1120 + 910 = 2030 kHz For low-side tuning we have fLO = fi − fIF = 1120 − 455 = 665 kHz fIM AGE = fi − 2fIF = 1120 − 910 = 210 kHz Problem 3.21 For high-side tuning fLO = fi + fIF = 1120 + 2500 = 3620 kHz fIM AGE = fi + 2fIF = 1120 + 5000 = 6120 kHz For low-side tuning fLO = fi − fIF = 1120 − 2500 = −1380 kHz fLO = 1380 kHz fIM AGE = fi − 2fIF = 1120 − 5000 = −3880 kHz fLO = 3880 kHz

CHAPTER 3. BASIC MODULATION TECHNIQUES

-1 0

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

-1 0 1

-1 0

Figure 3.7: In the preceding development for low-side tuning, recall that the spectra are symmetrical about f = 0. Problem 3.22 By deÞnition xc (t) = Ac cos [ω c t + kp m (t)] = Ac cos [ω c t + kp u(t − t0 )] The waveforms for the three values of kp are shown in Figure 3.7. The top pane is for kp = π, the middle pane is for kp = −π/2, and the bottom pane is for kp = π/4. Problem 3.23 Let φ (t) = β cos ω m t where ωm = 2πfm . This gives o n xc2 (t) = Ac Re ejω c t ejβ cos ωm t Expanding a Fourier series gives

ejβ cos ωm t =

∞ X

n=−∞

Cn ejnω m t

3.1. PROBLEMS

where ωm Cn = 2π

π/ωm

ejβ cos ω m t e−jnωm t dt

−π/ω m

With x = ω m t, the Fourier coeﬃcients become Z π 1 ejβ cos x e−jnx dx Cn = 2π −π ¢ ¡ Since cos x = sin x + π2 Z π π 1 ej [β sin(x+ 2 )−nx] dx Cn = 2π −π

With x = x + π2 , the preceding becomes 1 Cn = 2π

This gives Cn = e

j nπ 2

3π/2

ej [β sin y−ny+n 2 ] dy

−π/2

1 2π

j[β sin y−ny]

−π

where the limits have been adjusted by recognizing that the integrand is periodic with period 2π. Thus nπ Cn = ej 2 Jn (β) and

(

xc2 (t) = Ac Re ejωc t Taking the real part yields xc2 (t) = Ac

∞ X

Jn (β) ej (

nω m t+ nπ 2

n=−∞

)

h nπ i Jn (β) cos (ω c + nω m ) t + 2 n=−∞ ∞ X

The amplitude spectrum is therefore the same as in the preceding problem. The phase spectrum is the same as in the preceding problem except that nπ 2 is added to each term. Problem 3.24 Since sin(x) = cos(x − π2 ) we can write

which is

³ ´ π xc3 (t) = Ac sin(ω c t + β sin ωm t) = Ac cos ωc t − + β sin ω m t 2 o n xc3 (t) = Ac Re ej(ωc t−π/2) ej sin ωm t

CHAPTER 3. BASIC MODULATION TECHNIQUES

Since ej sin ωm t =

∞ X

Jn (β)ejnωm t

n=−∞

we have

(

j(ωc t−π/2)

xc3 (t) = Ac Re e Thus

∞ X

xc3 (t) = Ac

Jn (β)e

jnωm t

n=−∞

Thus

∞ X

)

o n Jn (β) Re ej(ωc t+nωm t−π/2)

h πi Jn (β) cos (ω c + nωm ) t − 2 n=−∞ ∞ X

Note that the amplitude spectrum of xc3 (t) is identical to the amplitude spectrum for both xc1 (t) and xc2 (t). The phase spectrum of xc3 (t) is formed from the spectrum of xc1 (t) by adding −π/2 to each term. For xc4 (t) we write ³ ´ π xc4 (t) = Ac sin(ω c t + β cos ωm t) = Ac cos ωc t − + β cos ω m t 2 Using the result of the preceding problem we write ( ) ∞ X j (nω m t+ nπ j(ωc t−π/2) ) 2 Jn (β) e xc4 (t) = Ac Re e n=−∞

This gives xc4 (t) = Ac

∞ X

n=−∞

Thus xc4 (t) = Ac

o n π nπ Jn (β) Re ej(ωc t+nω m t− 2 + 2 )

i h π Jn (β) cos (ω c + nωm ) t + (n − 1) 2 n=−∞ ∞ X

Compared to xc1 (t), xc2 (t) and xc3 (t), we see that the only diﬀerence is in the phase spectrum. Problem 3.25 From the problem statement xc (t) = Ac cos [2π (40) t + 10 sin (2π (5) t)]

3.1. PROBLEMS

The spectrum is determined using (3.101). Since fc = 40 and fm = 5 (it is important to note that fc is an integer multiple of fm ) there is considerable overlap of the portion of the spectrum centered about f = −40 with the portion of the spectrum centered about f = 40 for large β (such as 10). The terms in the spectral plot, normalized with respect to Ac , are given in Table 3.1 for f ≤ 0 (positive frequency terms and the term at dc). The terms resulting from the portion of the spectrum centered at f = 40 are denoted S40 and the terms resulting from the portion of the spectrum centered at f = −40 are denoted S−40 . Given the percision of the Table of Bessel coeﬃcients (Page 131 in the textbook), we have overlap for |f| ≤ 45. Where terms overlap, they must be summed. The sum is denoted ST in Table 3.1. In developing the Table 3.1 be sure to remember that J−n (β) = −Jn (β) for odd n. The power must now be determined in order to Þnd Ac . With the exception of the dc term, the power at each frequency is ST2 A2c /2 (the power for negative frequencies is equal to the power at positive frequencies and so the positive frequency power can simply be doubled to give the total power with the dc term excepted). The power at dc is ST2 A2c = (0.636)2 A2c . Carring out these operations with the aid of Table 3.1 gives the total power 0.8036A2c = 40 which is Ac =

40 = 7.0552 0.8036

The waveform xc (t) is illustrated in the top pane of Figure 3.8. The spectrum, normalized with respect to Ac , is illustrated in the bottom frame. The fact that xc (t) has a nonzero dc value is obvious. Problem 3.26 We are given J0 (3) = −0.2601 and J1 (3) = −0.3391. Using Jn+1 (β) =

2n Jn (β) − Jn−1 (β) β

with β = 3 we have 2 jn+1 (3) = nJn (3) − Jn−1 (3) 3 With n = 1, 2 J1 (3) − J0 (3) 3 2 = (0.3391) + 0.2601 = 0.4862 3

J2 (3) =

CHAPTER 3. BASIC MODULATION TECHNIQUES

Table 3.1: Data for Problem 3.25 f 125 120 115 110 105 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0

S40 0.001 0.002 0.005 0.012 0.029 0.063 0.123 0.207 0.292 0.318 0.219 −0.014 −0.234 −0.220 0.058 0.255 0.043 −0.246 −0.043 0.255 −0.058 −0.220 0.234 −0.014 −0.217 0.318

S−40 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.001 0.002 0.005 0.012 0.029 0.063 0.123 0.207 0.292 0.318

ST 0.001 0.002 0.005 0.012 0.029 0.063 0.123 0.207 0.292 0.318 0.219 −0.014 −0.234 −0.220 0.058 0.255 0.044 −0.244 −0.038 0.067 −0.029 −0.157 0.357 −0.193 0.075 0.636

3.1. PROBLEMS

x c (t)

0.5 0 -0.5 -1 0

0.02

0.04

0.06

0.08

0.1 Time

0.12

0.14

0.16

0.18

0.2

0.8

M agnitude

0.6 0.4 0.2 0 -150

-100

-50

0 Frequency

100

150

Figure 3.8: With n = 2, J3 (3) = =

4 J2 (3) − J1 (3) 3 4 (0.04862) + 0.3391 = 0.3091 3

Finally, with n = 3 we have J4 (3) = 2J3 (3) − J2 (3) = 2 (0.3091) + 0.4862 = 0.1320

Problem 3.27 The amplitude and phase spectra follow directly from the table of Fourier-Bessel coeﬃcients. The single-sided magnitude and phase spectra are shown in Figure 3.9. The magnitude spectrum is plotted assuming Ac = 1.

CHAPTER 3. BASIC MODULATION TECHNIQUES

0.4

M agnitude

0.3 0.2 0.1 0 600

700

800

900

1000 1100 Frequency

1200

1300

1400

700

800

900

1000 1100 Frequency

1200

1300

1400

P hase

-1 -2 -3 -4 600

Figure 3.9:

3.1. PROBLEMS

Problem 3.28 The modulated signal can be written i h xc (t) = Re 10 + 3ej2π(20)t + 5e−j2π(20)t ej2π(100)t

We will concentrate on the term in brackets, which is the complex envelope described in Chapter 2. Denoting the complex envelope by x ec (t), we can write x ec (t) = [10 + 3 cos 2π (20t) + 5 cos 2π (20) t] +j [3 sin 2π (20t) − 5 sin 2π (20) t] = [10 + 8 cos 2π (20t)] − j [2 sin 2π (20) t]

It follows from the deÞnition of xc (t) that x ec (t) = R (t) ejφ(t)

Thus

R2 (t) = [10 + 8 cos 2π (20t)]2 + 4 sin2 2π (20) t = 134 + 160 cos 2π (20) t + 30 cos 2π (40) t This gives R (t) = Also

p 134 + 160 cos 2π (20) t + 30 cos 2π (40) t φ (t) = tan−1

−2 sin 2π (20) t 10 + 8 cos 2π (20) t

Problem 3.29 ¡ ¢ Since sin x = cos x − π2 , we can write or where

h n io π xc (t) = Re ej200πt 5e−j40πt + 10 + 3ej (40πt− 2 ) © ª xc (t) = Re a (t) ej200πt a (t) = (5 cos 40πt + 10 + 3 sin 40πt) +j (5 sin 40πt − 3 cos 40πt)

CHAPTER 3. BASIC MODULATION TECHNIQUES

Thus if we write xc (t) as xc (t) = R (t) cos (200πt + φ (t)) the envelope, R (t), is given by i1 h 2 2 2 R (t) = (5 cos 40πt + 10 + 3 sin 40πt) + (5 sin 40πt − 3 cos 40πt)

and the phase deviation, φ (t), is given by φ (t) = tan−1

5 sin 40πt − 3 cos 40πt 10 + 5 cos 40πt + 3 sin 40πt

The envelope, R (t), can be simpliÞed and expressed in several diﬀerent forms. Problem 3.30 (a) Since the carrier frequency is 1000 Hertz, the general form of xc (t) is xc (t) = Ac cos [2π (1000) t + φ (t)] The phase deviation, φ (t), is therefore given by φ (t) = 20t2

rad

The frequency deviation is dφ = 40t dt or

rad/sec

1 dφ 20 = t 2π dt π

Hertz

(b) The phase deviation is φ (t) = 2π (500) t2 − 2π (1000) t

rad

and the frequency deviation is dφ dt

= 4π (500) t − 2π (1000) = 2000π (t − 1)

1 dφ = 1000 (t − 1) 2π dt

rad/sec

Hertz

3.1. PROBLEMS

rad

and the frequency deviation is dφ = 200π dt

rad/sec

1 dφ = 100 Hertz 2π dt which should be obvious from the expression for xc (t). (d) The phase deviation is √ φ (t) = 200πt + 10 t rad and the phase deviation is 1 dφ 5 1 = 200π + (10) t− 2 = 200π + √ dt 2 t

5 1 dφ = 100 + √ 2π dt 2π t

rad/sec

Hertz

Problem 3.31 (a) The phase deviation is φ(t) = 2π(30)

(8)dt = 480πt,

t≤4

The maximum phase deviation is φ(4) = 480π (4) = 1920π. The required plot is simply  t<0  0, φ(t) = 480πt, 0≤t<4  1920π, t≥4 (b) The frequency deviation, in Hz, is

0, 1 dφ = 30m(t) = 240, 2π dt 0

t<0 0≤t<4 t≥4

CHAPTER 3. BASIC MODULATION TECHNIQUES

The required sketch follows simply. (c) The peak frequency deviation is 8fd = 8 (30) = 240 Hertz (d) The peak phase deviation is Z 4 (8) dt = 2π (30) (8) (4) = 1920π 2π (30)

rad

The modulator output power is 1 1 P = A2c = (100)2 = 5000 2 2

Problem 3.32 (a) The message signal is

The phase deviation is

 0,    t − 4, m (t) = 8−t    0,

φ (t) = 2πfd

Watts

t<4 4≤t≤6 6≤t≤8 t>8

(t − 4) dt

¡ ¢ = 60π (−4) (t − 4) + 30π t2 − 16 ¢ ¡ = 30π t2 − 8t + 16 , 4 ≤ t ≤ 6 φ (t) = 2πfd

Also

(8 − t) dt + φ (6)

¡ ¢ = 120π + 60π (8) (t − 6) − 30π t2 − 36 ¡ ¢ = 120π − 30π t2 − 16t + 60 , 6 ≤ t ≤ 8 φ (t) = 240π, t > 8 φ (t) = 0, t < 4

The sketches follow immediately from the equations. (b) The frequency deviation in Hertz is 30m(t). (c) The peak phase deviation = 240π rad. (d) The peak frequency deviation = 120 Hertz.

3.1. PROBLEMS

(e) The carrier power is, assuming a suﬃciently high carrier frequency, 1 1 P = A2c = (100)2 = 5000 2 2

Watts

Problem 3.33 The frequency deviation in Hertz is the plot shown in Fig. 3.76 with the ordinate values multiplied by 25.The phase deviation in radians is given Z t Z t m (α) dα = 50π m (α) dα φ (t) = 2πfd For 0 ≤ t ≤ 1, we have

φ (t) = 50π

2αdα = 50πt2

For 1 ≤ t ≤ 2 φ (t) = φ(1) + 50π

¡ ¢ (5 − α) dα = 50π + 250π (t − 1) − 25π t2 − 1

= −175π + 250πt − 25πt2 For 2 ≤ t ≤ 3 φ (t) = φ(2) + 50π

3dα = 225π + 150π (t − 2)

For 3 ≤ t ≤ 4 φ (t) = φ(3) + 50π

2dα = 375π + 100π(t − 3)

Finally, for t > 4 we recognize that φ (t) = φ (4) = 475π. The required Þgure results by plotting these curves. Problem 3.34 The frequency deviation in Hertz is the plot shown in Fig. 3.77 with the ordinate values multiplied by 10. The phase deviation is given by Z t Z t m (α) dα = 20π m (α) dα φ (t) = 2πfd For 0 ≤ t ≤ 1, we have φ (t) = 20π

αdα = 10πt2

CHAPTER 3. BASIC MODULATION TECHNIQUES

For 1 ≤ t ≤ 2

t ¡ ¢ (α − 2) dα = 10π + 10π t2 − 1 − 40π(t − 1) φ (t) = φ(1) + 20π 1 ¢ ¡2 = 10π t − 4t + 4 = 10π (t − 2)2

For 2 ≤ t ≤ 4

φ (t) = φ(2) + 20π

¡ ¢ (6 − 2α)dα = 0 + 20π (6) (t − 2) − 20π t2 − 4

= −20π(t − 6t + 8)

Finally, for t > 4 we recognize that φ (t) = φ(4) = 0. The required Þgure follows by plotting these expressions. Problem 3.35 The frequency deviation in Hertz is the plot shown in Fig. 3.78 with the ordinate values multiplied by 5. The phase deviation in radians is given by φ (t) = 2πfd

m (α) dα = 10π

For 0 ≤ t ≤ 1, we have φ (t) = 10π

For 1 ≤ t ≤ 2 φ (t) = φ(1) + 10π

For 2 ≤ t ≤ 2.5 φ (t) = φ(2) + 10π

m (α) dα

(−2α)dα = −10πt2

2dα = −10π + 20π (t − 1) = 10π (2t − 3)

¡ ¢ (10 − 4α)dα = 10π + 10π (10) (t − 2) − 10π(2) t2 − 4

= 10π(−2t + 10t − 11) For 2.5 ≤ t ≤ 3

φ (t) = φ(2.5) − 10π

2.5

= 10π(−2t + 5.5)

2dα = 15π − 20π(t − 2.5)

3.1. PROBLEMS

For 3 ≤ t ≤ 4 φ (t) = φ(3) + 10π

(2α − 8) dα = 5π + 10π(t2 − 9) − 10π(8)(t − 3)

= 10π(t2 − 8t + 15.5) Finally, for t > 4 we recognize that φ (t) = φ(4) = −5π. The required Þgure follows by plotting these expressions. Problem 3.36 (a) The peak deviation is (12.5)(4) = 50 and fm = 10. Thus, the modulation index is 50 10 = 5. (b) The magnitude spectrum is a Fourier-Bessel spectrum with β = 5. The n = 0 term falls at 1000 Hz and the spacing between components is 10 Hz. The sketch is that of Figure 3.24 in the text. (c) Since β is not ¿ 1, this is not narrowband FM. The bandwidth exceeds 2fm . (d) For phase modulation, kp (4) = 5 or kp = 1.25. Problem 3.37 The results are given in the following table: Part a b c d

fd 20 200 2000 20000

Problem 3.38 From xc (t) = Ac

D = 5fd /W 0.004 0.04 0.4 4

∞ X

B = 2 (D + 1) W 50.2 kHz 52 kHz 70 kHz 250 kHz

Jn (β) cos (ω c + ω m ) t

n=−∞

we obtain

∞ 2 ® 1 2 X xc (t) = Ac J 2 (β) 2 n=−∞ n

We also know that (assuming that xc (t) does not have a signiÞcant dc component - see Problem 3.24) ® 2 ® 2 xc (t) = Ac cos2 [ω c t + φ (t)]

CHAPTER 3. BASIC MODULATION TECHNIQUES

which, assuming that ω c À 1 so that xc (t) has no dc component, is 2 ® 1 2 xc (t) = Ac 2

This gives

∞ 1 2 1 2 X 2 Ac = Ac J (β) 2 2 n=−∞ n

from which

∞ X

Jn2 (β) = 1

n=−∞

Problem 3.39 Since

1 dx = 2π

1 cos (β sin x − nx) dx + j 2π −π

1 Jn (β) = 2π

we can write 1 Jn (β) = 2π

−j(nx−β sin x)

−π

ej(β sin x−nx) dx

−π π

−π

sin (β sin x − nx) dx

The imaginary part of Jn (β) is zero, since the integrand is an odd function of x and the limits (−π, π) are even. Thus Z π 1 cos (β sin x − nx) dx Jn (β) = 2π −π Since the integrand is even Jn (β) =

1 π

cos (β sin x − nx) dx

which is the Þrst required result. With the change of variables λ = π − x, we have Z 1 π Jn (β) = cos [β sin (π − λ) − n (π − λ)] (−1) dλ π 0 Z 1 π = cos [β sin (π − λ) − nπ + nλ] dλ π 0 Since sin (π − λ) = sin λ, we can write Z 1 π Jn (β) = cos [β sin λ + nλ − nπ] dλ π 0

3.1. PROBLEMS

Using the identity cos (u − ν) = cos u cos ν + sin u sin ν with u = β sin λ + nλ and ν = nπ yields Jn (β) =

1 π +

cos [β sin λ + nλ] cos (nπ) dλ

1 π

sin [β sin λ + nλ] sin (nπ) dλ

Since sin (nπ) = 0 for all n, the second integral in the preceding expression is zero. Also cos (nπ) = (−1)n Thus

1 Jn (β) = (−1) π n

However

Thus

1 J−n (β) = π

cos [β sin λ + nλ] dλ

Jn (β) = (−1)n J−n (β) or equivalently J−n (β) = (−1)n Jn (β)

Problem 3.40 (a) Peak frequency deviation = 80 Hz (b) φ (t) = 8 sin (20πt) (c) β = 8 (d) Pi = 50 Watts, P0 = 16.76 Watts (e) The spectrum of the input signal is a Fourier_Bessel spectrum with β = 8. The n = 0 term is at the carrier frequency of 500 Hz and the spacing between components is 10 Hz. The output spectrum consistes of the n = 0 term and three terms each side of the n = 0 term. Thus the output spectrum has terms at 470, 480, 490, 500, 510, 520 and 530 Hz.

CHAPTER 3. BASIC MODULATION TECHNIQUES

Figure 3.10:

3.1. PROBLEMS

Problem 3.41 The required spectra are given in Figure 3.10. The modulation indices are, from top to bottom, β = 0.5, β = 1, β = 2, β = 5, and β = 10. Problem 3.42 We wish to Þnd k such that Pr = J02 (10) + 2

k X

n=1

J02 (10) ≥ 0.80

This gives k = 9, yielding a power ratio of Pr = 0.8747. The bandwidth is therefore B = 2kfm = 2 (9) (150) = 2700 Hz For Pr ≥ 0.9, we have k = 10 for a power ratio of 0.9603. This gives B = 2kfm = 2 (10) (150) = 3000 Hz

Problem 3.43 From the given data, we have fc1 = 110

kHz

fd1 = 0.05

This gives n=

fd2 = n (0.05) = 20

20 = 400 0.05

and fc1 = n (100)

kHz = 44 MHz

The two permissible local oscillator frequencies are f`0.1 = 100 − 44 = 56 MHz f`0.2 = 100 + 44 = 144 MHz The center frequency of the bandpass Þlter must be fc = 100 MHz and the bandwidth is ¡ ¢ B = 2 (D + 1) W = 2 (20 + 1) (10) 103 or

B = 420 kHz

CHAPTER 3. BASIC MODULATION TECHNIQUES

Problem 3.44 For the circuit shown H (f ) =

R E (f) = X (f ) R + j2πfL +

or H (f) = where

1 ³ 1 + j 2πfτ L −

1 j2πf C

1 2πf τ C

10−3 L = = 10−6 , 3 R 10 ¡ ¢¡ ¢ = RC = 103 10−9 = 10−6

τL = τC

A plot of the amplitude response shows that the linear region extends from approximately 54 kHz to118 kHz. Thus an appropriate carrier frequency is fc =

118 + 54 = 86 2

kHz

The slope of the operating characteristic at the operating point is measured from the amplitude response. The result is ¢ ¡ KD ∼ = 8 10−6 Problem 3.45 We can solve this problem by determining the peak of the amplitude response characteristic. This peak falls at 1 fp = √ 2π LC ¢ ¡ It is clear that fp > 100 MHz. Let fp = 150 MHz and let C = 0.001 10−12 . This gives L=

(2π) fp2 C

¢ ¡ = 1.126 10−3

We Þnd the value of R by trial and error using plots of the amplitude response. An appropriate value for R is found to be 1 MΩ. With these values, the discriminator constant is approximately ¢ ¡ KD ≈ 8.5 10−9 Problem 3.46

3.1. PROBLEMS

For Ai = Ac we can write, from (3.176), xr (t) = Ac [cos ω c t + cos (ω c + ω i ) t] which is xr (t) = Ac [(1 + cos ω i t) cos ω c t − sin ω i t sin ω c t] This yields xr (t) = R (t) cos [ω c t + ψ (t)] where −1

ψ (t) = tan This gives

· ¸ ¸ sin ω i t ωi t ωi t −1 tan = tan = 1 + cos ω i t 2 2

1 d yD (t) = 2π dt For Ai = −Ac , we get −1

ψ(t) = tan Since

we have Thus

2πfi t 2

1 = fi 2

¸ ¸ · − sin ω i t sin ω i t −1 = − tan 1 − cos ω i t 1 + cos ω i t

³π x´ sin x x = cot = tan − 1 − cos x 2 2 2

h ³ π x ´i x π − = − ψ(t) = tan−1 − tan 2 2 2 2 µ ¶ 1 d 2πfi t π 1 yD (t) = − = fi 2π dt 2 2 2

Finally, for Ai À Ac we see from the phasor diagram that ψ (t) ≈ θ (t) = ω i t and yD (t) =

KD d (2πfi t) = fi 2π dt

Problem 3.47 From Example 3.5, m (t) = Au (t). Thus Z t Au (α) dα = Akf t, φ (t) = kf

t≥0

CHAPTER 3. BASIC MODULATION TECHNIQUES

φ (t ) θ (t ) ψ (t )

Ak f KT

t 0

Figure 3.11: Also Θ (s) =

AKT kf 2 s (s + KT )

Taking the inverse transform gives θ (t) = Akf

µ ¶ 1 1 −KT t e − t+ u (t) KT KT

The phase error is ψ (t) = φ (t) − θ (t). This gives ψ (t) =

¢ Akf ¡ 1 − e−KT t u (t) KT

The maximum phase error occurs as t → ∞ and is clearly Akf /KT . Thus we require 0.2 =

Akf KT

and KT = 5Akf The VCO constant is contained in KT . The required sketch follows. Problem 3.48 For m (t) = A cos ω m t and φ (t) = Akf

cos ω m αdα =

Akf sin ω m t ωm

3.1. PROBLEMS

and Φ (s) =

Akf + ω 2m

The VCO output phase is Θ (s) = Φ (s)

Akf KT KT = s + KT (s + KT ) (s2 + ω 2m )

Using partial fraction expansion,Θ (s) can be expressed as · ¸ Akf KT 1 s KT Θ (s) = − + KT 2 + ω 2m s + KT s2 + ω 2m s2 + ω2m This gives, for t ≥ 0

· ¸ Akf KT KT −KT t θ (t) = e − cos ω m t + sin ω m t KT 2 + ω 2m ωm

The Þrst term is the transient response. For large KT , only the third term is signiÞcant. Thus, Ak K 2 1 dθ ¡ 2f T ¢ cos ω m t = eν (t) = Kν dt Kν KT + ω 2m Also, since KT is large, KT2 + ω 2m ≈ KT2 . This gives eν (t) ≈

Akf cos ω m t Kν

and we see that eν (t) is proportional to m (t). If kf = Kν , we have eν (t) ≈ m (t)

Problem 3.49 The Costas PLL is shown in Figure 3.57. The output of the top multiplier is m (t) cos ωc t [2 cos (ω c t + θ)] = m (t) cos θ + m (t) cos (2ω c t + θ) which, after lowpass Þltering, is m (t) cos θ. The quadrature multiplier output is m (t) cos ω c t [2 sin (ω c t + θ)] = m (t) sin θ + m (t) sin (2ω c t + θ) which, after lowpass Þltering, is m (t) sin θ. The multiplication of the lowpass Þlter outputs is m (t) cos θm (t) sin θ = m2 (t) sin 2θ

CHAPTER 3. BASIC MODULATION TECHNIQUES

dψ / dt

ψ A

Figure 3.12: Phase Detector

Loop Filter and Amplifier

VCO

ev ( t ) Bandpass Filter

e0 ( t )

Figure 3.13: as indicated. Note that with the assumed input m (t) cos ω c t and VCO output 2 cos (ω c t + θ), the phase error is θ. Thus the VCO is deÞned by dθ dψ = = −Kν eν (t) dt dt This is shown below. Since the dψ dt intersection is on a portion of the curve with negative slope, the point A at the origin is a stable operating point. Thus the loop locks with zero phase error and zero frequency error. Problem 3.50 ¡ ¢ With x (t) = A2πf0 t, we desire e0 (t) = A cos 2π 73 f0 t. Assume that the VCO output is a pulse train with frequency 13 fo . The pulse should be narrow so that the seventh harmonic is relatively large. The spectrum of the VCO output consists of components separated by

3.1. PROBLEMS

This component (the fundamental) tracks the input signal

Bandpass filter passband

1 f0 3

7 f0 3

Figure 3.14: 1 3 f0

with an envelope of sinc(τ f), where τ is the pulse width. The center frequency of the bandpass Þlter is 73 f0 and the bandwidth is on the order of 13 f0 as shown. Problem 3.51 The phase plane is deÞned by ψ = ∆ω − Kt sin ψ (t) at ψ = 0, ψ = ψ ss , the steady-state phase error. Thus µ ¶ µ ¶ ∆ω −1 ∆ω −1 ψss = sin = sin Kt 2π (100) For ∆ω = 2π (30) −1

ψ ss = sin For ∆ω = 2π (50)

−1

ψ ss = sin For ∆ω = 2π (80)

−1

ψ ss = sin

30 100

50 100

80 100

= 17.46 degrees

= 30 degrees

= 53.13 defrees

CHAPTER 3. BASIC MODULATION TECHNIQUES

For ∆ω = −2π (80)

−1

ψ ss = sin

−80 100

= −53.13 degrees

For ∆ω = 2π (120), there is no stable operating point and the frequency error and the phase error oscillate (PLL slips cycles continually). Problem 3.52 From (3.228) Kt

which is

s+a s+ε

Kt F (s) Θ (s) ³ ´ = = Φ (s) s + Kt F (s) s + Kt s+a s+ε Kt (s + a) Kt (s + a) Θ (s) = = 2 Φ (s) s (s + ε) + Kt (s + a) s + (Kt + ε) s + Kt a

Therefore s2 + 2ζω n s + ω 2n = s2 + (Kt + ε) s + Kt a This gives ωn = and

Problem 3.53 Since ω n = 2π (100) we have or Since

we have

p Kt a

Kt + ε ζ= √ 2 Kt a

p Kt a = 2π (100) ¡ ¢ Kt a = 4π2 104

1.1Kt Kt + ε 1 = ς=√ = √ 2 (2π) (100) 2 K a 2 t √ 2 (2π) (100) = 807.8 Kt = 1.1

Thus ε = 0.1Kt = 80.78

3.1. PROBLEMS

x PWM (t )

(⋅)dt

xPAM ( t )

x1 (t )

x2 ( t ) Pulse train

Figure 3.15: and

¡ ¢ 4π2 104 a= = 488.7 807.8

Problem 3.54 1 KD Kν . The VCO constant From (3.247), we see that the phase deviation is reduced by 1+ 2π is 25 Hz /volt. Thus rad/s/volt Kν = 2π (25) we may then write 5 D1 1 = = 12.5 = 1 + KD (2π) (25) D2 0.4 2π which gives 1 + 25KD = 12.5 Thus KD = 0.46

Problem 3.55 A system converting PWM to PAM can be realized as illustrated in Figure 3.15. The operation should be clear from the waveforms shown in Figure 3.16. The integrator can be realized by using a capacitor since, for a capacitor, Z 1 i (t) dt ν (t) = C Multiplication by the pulse train can be realized by sampling the capacitor voltage. Thus, the simple circuit is as illustrated in Figure 3.17.

CHAPTER 3. BASIC MODULATION TECHNIQUES

x PWM (t )

t 0

x1 (t )

t 0 x2 ( t )

t 0 x PAM (t )

t 0

Figure 3.16:

3.1. PROBLEMS

xPWM ( t ) = i ( t )

xPAM ( t ) sampling switch is closed for

nT − τ < t < nT

switch closes at t = nT to restart integration

Figure 3.17: Problem 3.56 Let A be the peak-to-peak value of the data signal. The peak error is 0.5% and the peakto-peak error is 0.01 A. The required number of quantizating levels is A = 100 ≤ 2n = q 0.01A so we choose q = 128 and n = 7. The bandwidth is B = 2W k log2 q = 2W k(7) The value of k is estimated by assuming that the speech is sampled at the Nyquist rate. Then the sampling frequency is fs = 2W = 8 kHz. Each sample is encoded into n = 7 pulses. Let each pulse be τ with corresponding bandwidth τ1 . For our case τ= Thus the bandwidth is

1 1 = nfs 2W n

1 = 2W n = 2W log2 q τ

and so k = 1. For k = 1 B = 2 (8, 000) (7) = 112 kHz

Problem 3.57

CHAPTER 3. BASIC MODULATION TECHNIQUES

Let the maximum error be λA where A is the peak-to-peak message signal. The peak-topeak error is 2λA and the minimum number of quantizing levels qmin = The wordlength is given by

A 1 = 2λA 2λ

¸ · 1 n = log2 2λ

where [x] is the smallest integer greater than or equal to x. With k = 1 (as in the previous problem), this gives a normalized bandwidth of · ¸ 1 B = log2 BN = 2W 2λ We make the following table λ 0.001 0.005 0.01 0.05 0.1 0.2 0.4 0.5

¡1¢ log 2λ 8.966 6.644 5.644 3.322 2.322 1.322 0.322 0

BN 9 7 6 4 3 2 1 0

A plot of BN as a function of λ gives the required plot. Problem 3.58 The message signal is m(t) = 4 sin 2π(10)t + 5 sin 2π(20)t The derivative of the message signal is dm (t) = 80π cos 2π (10) t + 200π cos 2π (20t) dt The maximum value of dm (t) /dt is obviously 280π and the maximum occurs at t = 0. Thus δ0 ≥ 280π Ts or 280π 280π = 5600 fs ≥ = δ0 0.05π

3.1. PROBLEMS

x1 (BW = W )

x2 (BW = W ) x3 (BW = 2W )

x4 (BW = 4W )

x5 (BW = 4W )

Output

Figure 3.18: Thus, the minimum sampling frequency is 5600 Hz. Problem 3.59 One possible commutator conÞguration is illustrated in Figure 3.18. The signal at the point labeled “output” is the baseband signal. The minimum commutator speed is 2W revolutions per second. For simplicity the commutator is laid out in a straight line. Thus, the illustration should be viewed as it would appear wrapped around a cylinder. After taking the sample at point 12 the commutator moves to point 1. On each revolution, the commutator collects 4 samples of x4 and x5 , 2 samples of x3 , and one sample of x1 and x2 . The minimum transmission bandwidth is X B = Wi = W + W + 2W + 4W + 4W i

= 12W

Problem 3.60 The single-sided spectrum for x (t) is shown in Figure 3.19. From the deÞnition of y (t) we have Y (s) = a1 X (f) + a2 X (f) ∗ X (f)

CHAPTER 3. BASIC MODULATION TECHNIQUES

Figure 3.19: The spectrum for Y (f ) is given in Figure 3.20. Demodulation can be a problem since it may be difficult to Þlter the desired signals from the harmonic and intermodulation distortion caused by the nonlinearity. As more signals are included in x (t), the problem becomes more difficult. The difficulty with harmonically related carriers is that portions of the spectrum of Y (f) are sure to overlap. For example, assume that f2 = 2f1 . For this case, the harmonic distortion arising from the spectrum centered about f1 falls exactly on top of the spectrum centered about f2 .

3.1. PROBLEMS

Y(f) a1

2a2W

a2W f f 2 − f1

Figure 3.20:

2 f2

f1 + f 2

2 f2